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我对 R psych 包中加权 kappa 的行为感到有些困惑。我有五个类(1、2、3、4、5)和三个注释器的数据。似乎每当一个特定的类(比如 1)只被一个注释器使用(导致该类的一致性矩阵的对角线上的空值)时,心理抱怨:

library(psych); 
df2 <- read.table('classes/WB/mcrae.quants.features.time.numeric', sep='\t'); 
cohen.weights <- matrix(c(0,5,35,95,100,5,0,30,90,95,35,5,0,60,65,95,90,60,0,5,100,95,65,5,0),ncol=5); 
cohen.kappa(df2,cohen.weights)
Error in `rownames<-`(`*tmp*`, value = c("2", "3", "4", "5")) : 
length of 'dimnames' [1] not equal to array extent
Calls: cohen.kappa -> cohen.kappa1 -> rownames<-
Execution halted

我的数据如下所示:

5       3       3
5       3       3
5       4       3
5       4       3
3       3       3
5       4       4
3       4       4
3       3       3
2       3       2
...
1       3       4

有人可以告诉我我做错了什么吗?当所有类都被至少两个注释器使用时,上面的代码才有效。谢谢!

dput(df2) 的输出:

structure(list(V1 = c(5L, 5L, 5L, 5L, 3L, 5L, 3L, 3L, 2L, 4L, 
3L, 4L, 3L, 4L, 5L, 3L, 5L, 5L, 3L, 5L, 3L, 5L, 3L, 5L, 4L, 5L, 
5L, 5L, 5L, 3L, 4L, 4L, 5L, 5L, 5L, 3L, 3L, 5L, 5L, 5L, 5L, 3L, 
3L, 5L, 5L, 3L, 4L, 3L, 4L, 3L, 5L, 5L, 4L, 3L, 5L, 5L, 5L, 5L, 
5L, 3L, 3L, 5L, 5L, 5L, 5L, 5L, 4L, 3L, 5L, 5L, 5L, 5L, 4L, 4L, 
4L, 5L, 4L, 4L, 3L, 3L, 5L, 3L, 4L, 3L, 5L, 1L), V2 = c(3L, 3L, 
4L, 4L, 3L, 4L, 4L, 3L, 3L, 4L, 4L, 4L, 3L, 3L, 3L, 3L, 4L, 4L, 
3L, 5L, 4L, 4L, 5L, 5L, 3L, 3L, 3L, 4L, 4L, 3L, 4L, 3L, 3L, 4L, 
4L, 3L, 4L, 5L, 4L, 4L, 3L, 4L, 3L, 3L, 4L, 3L, 4L, 3L, 4L, 4L, 
5L, 4L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 3L, 4L, 4L, 
4L, 4L, 4L, 4L, 5L, 3L, 4L, 4L, 3L, 4L, 4L, 3L, 3L, 3L, 4L, 3L, 
3L, 3L, 4L, 3L), V3 = c(3L, 3L, 3L, 3L, 3L, 4L, 4L, 3L, 2L, 3L, 
2L, 3L, 3L, 3L, 4L, 2L, 3L, 4L, 3L, 3L, 4L, 4L, 5L, 5L, 4L, 3L, 
3L, 3L, 5L, 3L, 3L, 4L, 4L, 4L, 3L, 3L, 4L, 4L, 4L, 4L, 3L, 4L, 
4L, 4L, 4L, 4L, 3L, 5L, 4L, 3L, 5L, 4L, 3L, 4L, 5L, 5L, 4L, 3L, 
5L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 3L, 3L, 4L, 4L, 3L, 3L, 4L, 4L, 
3L, 4L, 4L, 4L, 3L, 3L, 4L, 3L, 3L, 3L, 3L, 4L)), .Names = c("V1", 
"V2", "V3"), class = "data.frame", row.names = c(NA, -86L))
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1 回答 1

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需要变异性是因为 Kappa 会计算评估者间的可靠性并纠正机会一致性。有两个未知数,没有可变性,这是无法计算的。

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于 2015-03-26T14:46:44.227 回答